Question

好的，我的代码包含for循环中的嵌套查询

var query = records.find({$or:[{starter:data},{receiver:data}]},{});//check the records table for all persons the logged in user has spoken to
    query.sort('-createDate').exec(function (err, docs){
        if(err) throw err;

        for(var i=docs.length-1; i>= 0; i--)
        {

           var starter  = docs[i].starter;
            var receiver = docs[i].receiver;
            var lasttxt = docs[i].lastMessage; 

            if (starter == socket.usernames){
              var target = receiver;
            }else
            {
              var target = starter;
            }

          usersrec.find({username:target},{}).lean().exec(function (errx, docx){
                if(errx) throw errx;

                docx[0].message = lasttxt;
                socket.emit('usernames', docx);
          });
        }
    })

它的意思是获取当前登录用户所说的每个人的最后一条消息并存储在lasttxt变量中。问题是它只获取数据库中最后一个用户的最后一条消息然后它将最后一条消息分配给每个人作为他们自己的最后一个消息。

这不会影响数据库的记录。只是客户端 我错过了什么？

Answer 1

为了导航js异步，我使用socket.io进行了一些来回发布，并且它工作了

服务器端的

var query = records.find({$or:[{starter:data},{receiver:data}]},{});//check the records table for all persons the logged in user has spoken to
query.sort('-createDate').exec(function (err, docs){
    if(err) throw err;

    for(var i=docs.length-1; i>= 0; i--)
    {

       var starter  = docs[i].starter;
        var receiver = docs[i].receiver;
        var lasttxt = docs[i].lastMessage; 

        if (starter == socket.usernames){
          var target = receiver;
        }else
        {
          var target = starter;
        }

      var userlast = target+" "+lasttxt;
                socket.emit('lastly', userlast);//Emit the username and last message for the client to emit back here
    }
})

在您的客户端，选择发出的数据

 socket.on('lastly', function(data){//Recieve the data and send right back
                  socket.emit('lastly2', data);
              });

回到服务器端，拿起发回的数据

socket.on('lastly2', function(data){//receive the username and last message to work with

var check = data;
var space = check.indexOf(' ');
var name = check.substr(0, space);
var msg = check.substr(space+1);

usersrec.find({username:name},{}).lean().exec(function (errx, docx){
            if(errx) throw errx;

            docx[0].message = msg;
            socket.emit('usernames', docx);
      });

是的，它可能是非正统的，但至少它完成了工作。我愿意接受更好的建议

MongoDB嵌套查询仅返回上次发生的结果

1 个答案: